QBManager

COORDINATION COMPOUNDS

274152 Among ${[Fe {(H_{2} O)}_{6}]}^{3 +}, {[Fe (CN)_{6}]}^{3 -}, {[Fe (Cl)_{6}]}^{3 -}$ species, the hybridization state of the $Fe$ atom are, respectively.

1

d^{2} {sp}^{3}, d^{2} {sp}^{3}, {sp}^{3} d^{2}

2

{sp}^{3} d^{2}, d^{2} {sp}^{3}, d^{2} {sp}^{3}

3

{sp}^{3} d^{2}, d^{2} {sp}^{3}, {sp}^{3} d^{2}

4 None of the above

VITEEE-2014

COORDINATION COMPOUNDS

274154 The number of unpaired electrons calculated in ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[Co (F_{6})]}^{3 -}$

1 4 and 4

2 0 and 2

3 2 and 4

4 0 and 4

Explanation:

(D) : In ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[Co (F_{6})]}^{3 -}$ both the oxidation state of cobalt ion is +3 . In first case ${NH}_{3}$ is the neutral ligand which is a strong field ligand.
Hence, the electrons in Co $(+ 3)$ i.e. $4 s^{0} 3 d^{6}$ get paired to form inner orbital complex.
Hence, no unpaired electron.
On the other hand $F^{-}$ is a weak field ligand hence it forms an outer orbital complex with 4 unpaired electrons.

VITEEE- 2008

COORDINATION COMPOUNDS

274155 In ${TeCl}_{4}$ the central atom tellurium involves

1

{sp}^{3}

hybridization

2

{sp}^{3} d

hybridization

3

{sp}^{3} d^{2}

hybridization

4

{dsp}^{2}

hybridization

Explanation:

(B) : Hybridisation $= \frac{1}{2} [$ Number of valence electrons of central atom + no. of monovalent atoms attached to it + negative charge if any - positive charge if any]
$= \frac{1}{2} [6 + 4 + 0 - 0] = 5 = {sp}^{3} d$
$[∵ Te (52) = [Kr] 4 d^{10} 5 s^{2} 5 p^{4}]$

VITEEE- 2007

COORDINATION COMPOUNDS

274156 The hybridization of nickel in nickel tetracarbonyl is

1

{sp}^{3}

2

{sp}^{2}

3

{dsp}^{2}

4

d^{2} s^{3}

Explanation:

(A) : The chemical formula of nickeltetracarbonyl is $[Ni (CO)_{4}]$ . The oxidation number of the complex is 0 .
$_{28} Ni = [Ar] 3 d^{8} 4 s^{2}$
original image

COMEDK 2011

COORDINATION COMPOUNDS

274152 Among ${[Fe {(H_{2} O)}_{6}]}^{3 +}, {[Fe (CN)_{6}]}^{3 -}, {[Fe (Cl)_{6}]}^{3 -}$ species, the hybridization state of the $Fe$ atom are, respectively.

1

d^{2} {sp}^{3}, d^{2} {sp}^{3}, {sp}^{3} d^{2}

2

{sp}^{3} d^{2}, d^{2} {sp}^{3}, d^{2} {sp}^{3}

3

{sp}^{3} d^{2}, d^{2} {sp}^{3}, {sp}^{3} d^{2}

4 None of the above

VITEEE-2014

COORDINATION COMPOUNDS

274154 The number of unpaired electrons calculated in ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[Co (F_{6})]}^{3 -}$

1 4 and 4

2 0 and 2

3 2 and 4

4 0 and 4

Explanation:

(D) : In ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[Co (F_{6})]}^{3 -}$ both the oxidation state of cobalt ion is +3 . In first case ${NH}_{3}$ is the neutral ligand which is a strong field ligand.
Hence, the electrons in Co $(+ 3)$ i.e. $4 s^{0} 3 d^{6}$ get paired to form inner orbital complex.
Hence, no unpaired electron.
On the other hand $F^{-}$ is a weak field ligand hence it forms an outer orbital complex with 4 unpaired electrons.

VITEEE- 2008

COORDINATION COMPOUNDS

274155 In ${TeCl}_{4}$ the central atom tellurium involves

1

{sp}^{3}

hybridization

2

{sp}^{3} d

hybridization

3

{sp}^{3} d^{2}

hybridization

4

{dsp}^{2}

hybridization

Explanation:

(B) : Hybridisation $= \frac{1}{2} [$ Number of valence electrons of central atom + no. of monovalent atoms attached to it + negative charge if any - positive charge if any]
$= \frac{1}{2} [6 + 4 + 0 - 0] = 5 = {sp}^{3} d$
$[∵ Te (52) = [Kr] 4 d^{10} 5 s^{2} 5 p^{4}]$

VITEEE- 2007

COORDINATION COMPOUNDS

274156 The hybridization of nickel in nickel tetracarbonyl is

1

{sp}^{3}

2

{sp}^{2}

3

{dsp}^{2}

4

d^{2} s^{3}

Explanation:

(A) : The chemical formula of nickeltetracarbonyl is $[Ni (CO)_{4}]$ . The oxidation number of the complex is 0 .
$_{28} Ni = [Ar] 3 d^{8} 4 s^{2}$
original image

COMEDK 2011

COORDINATION COMPOUNDS

274152 Among ${[Fe {(H_{2} O)}_{6}]}^{3 +}, {[Fe (CN)_{6}]}^{3 -}, {[Fe (Cl)_{6}]}^{3 -}$ species, the hybridization state of the $Fe$ atom are, respectively.

1

d^{2} {sp}^{3}, d^{2} {sp}^{3}, {sp}^{3} d^{2}

2

{sp}^{3} d^{2}, d^{2} {sp}^{3}, d^{2} {sp}^{3}

3

{sp}^{3} d^{2}, d^{2} {sp}^{3}, {sp}^{3} d^{2}

4 None of the above

VITEEE-2014

COORDINATION COMPOUNDS

274154 The number of unpaired electrons calculated in ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[Co (F_{6})]}^{3 -}$

1 4 and 4

2 0 and 2

3 2 and 4

4 0 and 4

Explanation:

(D) : In ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[Co (F_{6})]}^{3 -}$ both the oxidation state of cobalt ion is +3 . In first case ${NH}_{3}$ is the neutral ligand which is a strong field ligand.
Hence, the electrons in Co $(+ 3)$ i.e. $4 s^{0} 3 d^{6}$ get paired to form inner orbital complex.
Hence, no unpaired electron.
On the other hand $F^{-}$ is a weak field ligand hence it forms an outer orbital complex with 4 unpaired electrons.

VITEEE- 2008

COORDINATION COMPOUNDS

274155 In ${TeCl}_{4}$ the central atom tellurium involves

1

{sp}^{3}

hybridization

2

{sp}^{3} d

hybridization

3

{sp}^{3} d^{2}

hybridization

4

{dsp}^{2}

hybridization

Explanation:

(B) : Hybridisation $= \frac{1}{2} [$ Number of valence electrons of central atom + no. of monovalent atoms attached to it + negative charge if any - positive charge if any]
$= \frac{1}{2} [6 + 4 + 0 - 0] = 5 = {sp}^{3} d$
$[∵ Te (52) = [Kr] 4 d^{10} 5 s^{2} 5 p^{4}]$

VITEEE- 2007

COORDINATION COMPOUNDS

274156 The hybridization of nickel in nickel tetracarbonyl is

1

{sp}^{3}

2

{sp}^{2}

3

{dsp}^{2}

4

d^{2} s^{3}

Explanation:

(A) : The chemical formula of nickeltetracarbonyl is $[Ni (CO)_{4}]$ . The oxidation number of the complex is 0 .
$_{28} Ni = [Ar] 3 d^{8} 4 s^{2}$
original image

COMEDK 2011

COORDINATION COMPOUNDS

274152 Among ${[Fe {(H_{2} O)}_{6}]}^{3 +}, {[Fe (CN)_{6}]}^{3 -}, {[Fe (Cl)_{6}]}^{3 -}$ species, the hybridization state of the $Fe$ atom are, respectively.

1

d^{2} {sp}^{3}, d^{2} {sp}^{3}, {sp}^{3} d^{2}

2

{sp}^{3} d^{2}, d^{2} {sp}^{3}, d^{2} {sp}^{3}

3

{sp}^{3} d^{2}, d^{2} {sp}^{3}, {sp}^{3} d^{2}

4 None of the above

VITEEE-2014

COORDINATION COMPOUNDS

274154 The number of unpaired electrons calculated in ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[Co (F_{6})]}^{3 -}$

1 4 and 4

2 0 and 2

3 2 and 4

4 0 and 4

Explanation:

(D) : In ${[Co {({NH}_{3})}_{6}]}^{3 +}$ and ${[Co (F_{6})]}^{3 -}$ both the oxidation state of cobalt ion is +3 . In first case ${NH}_{3}$ is the neutral ligand which is a strong field ligand.
Hence, the electrons in Co $(+ 3)$ i.e. $4 s^{0} 3 d^{6}$ get paired to form inner orbital complex.
Hence, no unpaired electron.
On the other hand $F^{-}$ is a weak field ligand hence it forms an outer orbital complex with 4 unpaired electrons.

VITEEE- 2008

COORDINATION COMPOUNDS

274155 In ${TeCl}_{4}$ the central atom tellurium involves

1

{sp}^{3}

hybridization

2

{sp}^{3} d

hybridization

3

{sp}^{3} d^{2}

hybridization

4

{dsp}^{2}

hybridization

Explanation:

(B) : Hybridisation $= \frac{1}{2} [$ Number of valence electrons of central atom + no. of monovalent atoms attached to it + negative charge if any - positive charge if any]
$= \frac{1}{2} [6 + 4 + 0 - 0] = 5 = {sp}^{3} d$
$[∵ Te (52) = [Kr] 4 d^{10} 5 s^{2} 5 p^{4}]$

VITEEE- 2007

COORDINATION COMPOUNDS

274156 The hybridization of nickel in nickel tetracarbonyl is

1

{sp}^{3}

2

{sp}^{2}

3

{dsp}^{2}

4

d^{2} s^{3}

Explanation:

(A) : The chemical formula of nickeltetracarbonyl is $[Ni (CO)_{4}]$ . The oxidation number of the complex is 0 .
$_{28} Ni = [Ar] 3 d^{8} 4 s^{2}$
original image

COMEDK 2011

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